Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<long long> tri(100005);
bool check(long long x) {
  int low = 1, high = 100000, mid = 0;
  while (low <= high) {
    mid = low + (high - low) / 2;
    if (tri[mid] > x) {
      high = mid - 1;
    } else if (tri[mid] < x) {
      low = mid + 1;
    } else {
      return true;
    }
  }
  return false;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < 100005; i++) {
    tri[i] = (i * (i + 1)) / 2LL;
  }
  for (int i = 1; i <= 100000; i++) {
    long long x = n - tri[i];
    if (x <= 0) {
      cout << "NO\n";
      return;
    }
    if (check(x)) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}